Physics Factsheet
www.curriculumpress.co.uk
Number 66
Forces on Vehicles
This Factsheet focuses on applying concepts from mechanics to motor
vehicles. It covers:
• how cars move
• how cars brake
• power
• cars in motion
How cars brake
Braking provides a torque opposite to that of the engine. This is due to
friction. There are two types of brakes:
Disc brakes
brake pads hydraulic caliper
A later Factsheet will deal with stopping distances, accidents and car
safety features.
wheel •
disc
•
Before working through this Factsheet you should understand
• the ideas of balanced and unbalanced forces
• that two bodies interacting exert equal and opposite forces on each
other
• the relationship between work, energy and power (Factsheet 05)
axle
disc
Drum brakes
hydraulic
cylinder
How cars move
•
•
•
force of
tyre on road
force of
road on tyre
•
•
brake
shoes
The car engine exerts a torque on the
axle of the wheel
The wheel turns and exerts a
backward force on the road, where
the tyre touches it
The road exerts an equal forward
force on the tyre - due to friction.
The unbalanced forward force of the
road on the tyre pushes the car
forward.
power =
work done
time taken
We also know that:
work done by a force = force × distance moved in direction of force
We can put these two equations together to find an expression for the
power of a car's engine:
power =
Friction is required to push the car forwards.
motive force × distance moved
time taken
Rearranging:
power = motive force ×
What affects the motive force?
The motive force is affected both by factors due to the particular car
and by external factors:
•
•
•
•
a metal drum is fixed inside the
wheel
braking is carried out by pushing
the brake shoes against the drum
friction slows the car down
Power
Power is the rate of doing work - if work is being done at a constant rate,
then we have
direction of
motion of the car
•
•
drum
The force driving the car forward is called the motive force
torque provided by engine
•
a metal disc is attached to the axle
the wheel is on
braking is carried out by pushing
the brake pads against the disc
friction slows the car down
distance moved
time taken (s)
This gives:
P = Fv
The weather conditions - if it is icy or wet, friction is reduced, so
even if the engine is working as hard, there will be less friction between
the wheel and the road, and so the force pushing the car forward is
reduced
The road surface - again this affects the friction.
P = power (W)
F = motive force (N)
v = speed (ms-1)
Example
A car's engine has a power of 10kW.
The car is travelling at 72kmh-1
Find the motive force being exerted by the engine
The power of the engine - a more powerful engine produces a greater
motive force
The tyre treads - treads are designed to ensure a good grip on the road.
Water on the road should move up into the gaps in the treads; if the
tyres are in bad condition and are too smooth, the friction is reduced.
First convert 72kmh-1 to ms-1
72 kmh-1 = 72 000 mh-1
Use P = Fv
10 000 = F (20)
F = 500N
1
72 000 mh-1 = 72 000/3600 = 20 ms-1
Physics Factsheet
66 - Forces on Vehicles
Cars in Motion
The diagram below shows the main forces exerted on a car in motion
Car towing a trailer
Air resistance
Air resistance
Contact
force
Contact force
Contact force
Drag
Motive
force
•
•
T
Motive
force
Drag
Weight
T is the tension in the tow bar between the car and the trailer. It acts as
a backwards force on the car and a forward one on the trailer.
Again, vertical forces are balanced when travelling along a level road, and
horizontal forces are balanced if the car and trailer are moving at constant
velocity.
When tackling this sort of situation, there are two approaches:
• Consider the car and trailer as one system, with total mass the sum of
the two masses and total resistance the sum of both resistances. This
is a good approach to take if you do not need the tension in the tow bar.
The method is then exactly the same as considering the car on its own.
• Obtain two sets of equations from considering the car and trailer
separately. You need to do this if the tension in the tow bar is required.
Typical Exam Question
A car has mass 800kg. It is moving at a steady speed of 15 ms-1,
and the total drag forces on it are 500N.
(a) Find the power of the engine
Typical Exam Question
A car has of mass 900kg is towing a trailer of mass 300kg. The car's
engine has a power of 10.8kW. The total resistances to motion of the
car and trailer are 400N and 200N respectively, and these are
constant. The car and trailer are moving at constant velocity
(a) Find the tension in the tow bar
(b) Find the speed with which the car and trailer are moving.
The driver starts to accelerate at 1 ms-2.
(b) Find the motive force being exerted, assuming the total drag
forces are unchanged
(a) "Total drag forces" means the total of drag on the wheels and air
resistance.
Since the car is moving at a steady speed, the horizontal forces are
balanced
So: motive force = 500N
(a) Constant velocity, so forces are balanced
Considering trailer (it's simpler as there are fewer forces on it):
Tension - 200N = 0
Tension = 200N
Power = Fv
= 500 × 15
= 7500W (7.5 kW)
(b) The only equation we have with speed in it is the one for power.
Since we know the power of the engine, if we can find the motive
force, we'll be able to work out the speed.
(b) Using: resultant force = ma
resultant force = 800 ×1 = 800N
Considering car:
F - 400 - T = 0
F = 400 + T = 600N
So motive force - 500N = 800N
So motive force = 1300N
Exam Hint:- Many candidates lose marks through assuming that the
forces on a vehicle will always balance - always read through the
question to make sure it is moving at constant velocity before using this
Since P = Fv, 10800 = 600v
18ms-1 = v
Weight Transfer during braking or acceleration
The diagram shows the forces on a car when it is braking. For simplicity,
air resistance has been omitted, as it is relatively small.
Rr
Rf
W
Weight
Drag
Weight
The vertical forces on the car are balanced when the car is driving on a
level road, since it is not accelerating upwards or downwards.
If the car is moving at constant velocity (i.e. neither its speed nor
direction of motion are changing) then the horizontal forces are also
balanced.
If the car is accelerating (or decelerating), then the horizontal forces are
unbalanced, producing a resultant forwards (or backwards) force
respectively. The acceleration can be found using F = ma.
Br
Air
resistance
Contact force
T
•
Contact
force
D + Bf
Bf , Br = braking forces
at front and rear wheels
D = drag
Rf , Rr = contact forces
at front and rear wheels
W = weight of car
One consequence of this is that the friction between the tires and the road
changes - the size of the frictional force is proportional to the size of the
contact force, so the smaller the contact force, the less friction. So when
braking, the front tires have more grip on the road and the back tires less.
The opposite situation applies when accelerating forward. The motive
force has an anticlockwise moment about the centre of mass, which results
in the front of the car feeling lighter and the rear heavier - and the friction
between the front tires and the road being smaller.
The braking forces B1 and B2 have a clockwise moment about the centre of
mass of the car. This means that they will tend to turn the car clockwise so the front of the car tends to be pushed into the ground. This is felt by the
driver as the rear of the car getting lighter.
When a car goes round a bend, there is an unbalanced force in the direction
of the bend (eg to the left for a bend to the left). This also has a turning effect
- it tends to make the outside edge of the car feel comparatively heavier.
The car doesn't actually turn, so the contact forces Rf and Rr change to
compensate - Rf gets large, and since Rf and Rr add to W, Rr gets smaller.
Exam Hint:- Although this may not be explicitly on your specification,
it could be examined as a way of applying your knowledge of moments
2
Physics Factsheet
66 - Forces on Vehicles
Exam Workshop
Questions
This is a typical weak student’s answer to an exam question. The
comments explain what is wrong with the answers and how they can
be improved. The examiner’s answer is given below.
1. Explain how the engine of a car makes it move.
(a) Using a suitable diagram, explain how the engine of a car makes
it move
[4]
3. (a) Draw a diagram to show the forces acting on a car when it is in
motion.
(b) Explain the conditions under which any of these forces are
balanced.
2. State and explain four factors that affect the motive force on a car
wheel turns
this way
friction pushes the car
forward
4. The motive force of a car is 900N. It is travelling at 60 kmh-1.
Calculate the motive power of its engine.
2/4
Although the candidate seems to know how it works, there is not
enough detail here for four marks - s/he needed to mention that the
engine produced the rotation of the wheel, and that the friction was
the equal and opposite force to the push of the wheel on the ground.
5. A car is travelling at a constant velocity of 25 ms-1 on a flat straight
road. The total drag forces on the car are 500N. Find the power of
the engine.
6. A car has mass 600kg. When its engine is working at 18 kW, it is
travelling at 12 ms-1 and accelerating at 1.5 ms-2.
Calculate the total resistive forces acting on the car.
(b) State and explain the effect of water on the road on the motive
force of a car
[2]
Reduces it
1/2
7. A car of mass 900kg is towing a caravan of mass 800kg along a level
straight road at a constant velocity of 27 ms-1. The total drag forces
on the car and the caravan are 600N and 500N respectively.
Calculate
(a) the tension in the tow bar
(b) the power of the car's engine
The candidate has not read the question! S/he has just "stated" the
effect, not explained it.
(c) A car's engine has motive power 10kW. Calculate the motive
force exerted by the engine when the car is travelling at a constant
velocity of 20 ms-1
[2]
P = Fv
10 = F20
F = 0.5N
1/2
The right method - but units wrong! The candidate has read kW
as W. S/he should have realised 0.5N (around the weight of a
chocolate bar!) is not a sensible motive force.
(b) For car: F - 600 - T = 0
So F = 1100N
So P = Fv = 1100×27 = 29700W (29.7 kW)
(d) The car from (c) is logged travelling at 18 ms-1 along a level
straight road. Its engine is working at the same rate as before.
The forces resisting the motion are unchanged.
Explain why the car will accelerate
[2]
20ms-1 is the only steady speed at that power so it accelerates to
reach it.
0/2
7. (a) For caravan: T - 500 = 0
So T = 500N
F - resistive forces = 900N
1500 - resistive forces = 900N
resistive forces = 600N
Although in a sense the candidate is right - with unchanged resistive
forces, 20ms-1 is the only possible steady speed with the engine
working at 10kW - s/he has not demonstrated that s/he understands
the physics of why this is so - nor why acceleration is produced.
Using resultant force = ma:
resultant force = 600(1.5) = 900N
6. P = Fv
18000 = F12
F = 1500 N
(e) The car from (c) does work of 4.0× 105 J against a braking force
before coming to rest. Assuming the braking force is 1000N
calculate the distance travelled during braking.
[2]
2as = v2 - u2
Kinetic energy = 4.0× 105 = ½mv2
v2 = 4.0× 105 /m
1000 = ma
0/2
5. Constant velocity so forces balanced
F = 500N
P = Fv = 500(25) = 12500W (12.5 kW)
4. 60 kmh-1 = 60 000 mh-1 = 162/3 ms-1
P = Fv = 900× 162/3 = 15000W (15 kW)
Unfortunately the candidate is not using the correct formula s/he should have realised that the method was impossible without
the mass. When given work done and a force, you need to use
"Work done = force × distance moved
Answers
Answers to questions 1 - 3 can be found in the text.
Examiner’s Answer
(a) Diagram as on page 1.
Engine makes wheel rotate!. Wheel pushes against ground!. Force
of ground on wheel is equal !to force of wheel on ground, which
provides forwards! force.
(b) Reduces it !because motive force depends on friction, and a wet road
surface provides less friction!
(c) P = Fv 10000 = F20! F = 500N!
(d) Since F = P/v, F is larger!. Resistive forces unchanged so there is a
resultant forward force!
(e) Work done = F × d. 4.0× 105= 1000d!. d = 400m!
Acknowledgements: This Physics Factsheet was researched and written
by Cath Brown.
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ISSN 1351-5136
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